The eletron pairs repel each other in a tetrahedral shape.
<h3>
Answer:</h3>
7.4797 g AlF₃
<h3>
General Formulas and Concepts:</h3>
<u>Math</u>
<u>Pre-Algebra</u>
Order of Operations: BPEMDAS
- Brackets
- Parenthesis
- Exponents
- Multiplication
- Division
- Addition
- Subtraction
<u>Chemistry</u>
<u>Stoichiometry</u>
- Reading a Periodic Table
- Using Dimensional Analysis
<h3>
Explanation:</h3>
<u>Step 1: Define</u>
[RxN] 2AlF₃ + 3K₂O → 6KF + Al₂O₃
[Given] 15.524 g KF
<u>Step 2: Identify Conversions</u>
[RxN] 6 mol KF = 2 mol AlF₃
Molar Mass of K - 39.10 g/mol
Molar Mass of F - 19.00 g/mol
Molar Mass of Al - 26.98 g/mol
Molar Mass of KF - 39.10 + 19.00 = 58.1 g/mol
Molar Mass of AlF₃ - 26.98 + 3(19.00) = 83.98 g/mol
<u>Step 3: Stoichiometry</u>
- Set up:

- Multiply/Divide:

<u>Step 4: Check</u>
<em>Follow sig fig rules and round. We are given 5 sig figs.</em>
7.47966 g AlF₃ ≈ 7.4797 g AlF₃
A solution is the answer to a problem
Answer:
<u><em></em></u>
- <u><em>pOH = 0.36</em></u>
Explanation:
Both <em>potassium hydroxide</em> and <em>lithium hydroxide </em>solutions are strong bases, so you assume 100% dissociation.
<u>1. Potassium hydroxide solution, KOH</u>
- Volume, V = 304 mL = 0.304 liter
- number of moles, n = M × V = 0.36M × 0.304 liter = 0.10944 mol
- 1 mole of KOH produces 1 mol of OH⁻ ion, thus the number of moles of OH⁻ is 0.10944
<u>2. LIthium hydroxide, LiOH</u>
- Volume, V = 341 mL = 0.341 liter
- number of moles, n = M × V = 0.341 liter × 0.51 M = 0.17391 mol
- 1mole of LiOH produces 1 mol of OH⁻ ion, thus the number of moles of OH⁻ is 0.17391
<u />
<u>3. Resulting solution</u>
- Number of moles of OH⁻ ions = 0.10944 mol + 0.17391 mol = 0.28335 mol
- Volume of solution = 0.304 liter + 0.341 liter = 0.645 liter
- Molar concentration = 0.28335 mol / 0.645 liter = 0.4393 M
<u />
<u>4. </u><em><u>pOH</u></em>
← answer